Question: You have found the following ages (in years) of all 5 zebras at your local zoo: $ 15,\enspace 7,\enspace 14,\enspace 9,\enspace 14$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{15 + 7 + 14 + 9 + 14}{{5}} = {11.8\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $15$ years $3.2$ years $10.24$ years $^2$ $7$ years $-4.8$ years $23.04$ years $^2$ $14$ years $2.2$ years $4.84$ years $^2$ $9$ years $-2.8$ years $7.84$ years $^2$ $14$ years $2.2$ years $4.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{10.24} + {23.04} + {4.84} + {7.84} + {4.84}} {{5}} $ $ {\sigma^2} = \dfrac{{50.8}}{{5}} = {10.16\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{10.16\text{ years}^2}} = {3.2\text{ years}} $ The average zebra at the zoo is 11.8 years old. There is a standard deviation of 3.2 years.